If z1=(sqrt3-1)+(sqrt3+1)i and z2=-sqrt3+i find arg z1 and arg z2; hence, calculate arg(z1z2)?

1 Answer
Jan 16, 2018

#arg(z_1z_2)=(5pi)/4#

Explanation:

Modulus of a complex number is its absolute value i.e. if #z=a+ib#, #|z|=sqrt(a^2+b^2)# and

#argz=theta# if #z=|z|(costheta+isintheta)# and hence #argz=tan^(-1)(b/a)#

also note that using De Moivre's theorem #arg(z_1z_2)=argz_1+argz_2#

as if #z_1=r_1(costheta_1+isintheta_1)# and #z_2=r_2(costheta_2+isintheta_2)#

#z_1z_2=r_1r_2((costheta_1costheta_2-sintheta_1sintheta_2)+i(sintheta_1costheta_2-costheta_1sintheta_2))#

= #r_1r_2(cos(theta_1+theta_2)+isin(theta_1+theta_2))#

As #z_1=(sqrt3-1)+i(sqrt3+1)i#

now #|z_1|=sqrt((sqrt3-1)^2+(sqrt3+1)^2)#

and #argz_1=tan^(-1)

= #sqrt8=2sqrt2#, we get

#z_1=tan^(-1)((sqrt3+1)/(sqrt3-1))=(5pi)/12#

see details here

Similarly as #z_2=-sqrt3+i#, #argz_2=tan^(-1)(-1/sqrt3)=(5pi)/6#

Note #|z_2|=2#

Hence #arg(z_1z_2)=(5pi)/12+(5pi)/6=(5pi)/4#

Check #-># #((sqrt3-1)+i(sqrt3+1))(-sqrt3+i)#

= #-3+sqrt3+(sqrt3-1)i+i(-3-sqrt3)+i^2(sqrt3+1)#

= #-3+sqrt3+(sqrt3-1-3-sqrt3)i-sqrt3-1#

= #-4-4i#

= #4sqrt2(-1/sqrt2-1/sqrt2i)#

= #4sqrt2(cos((5pi)/4)+isin((5pi)/4))#

also #|z_1z_2|=4sqrt2=|z_1||z_2|#